Question

Find AB and BC from the following figure.​

Answer 1

Step-by-step explanation:

in the given fig . using trigonometry we find that

$\tan( {45}^{\degree} ) = \frac{ab}{bd} \\ \frac{ab}{bc + cd } = 1 \\ ab = bc + 20$

$and \:$

$\tan( {60}^{ \degree} ) = \frac{ab}{bc} \\ \sqrt{3} = \frac{ab}{bc} \\ ab = \sqrt{3} bc$

comparing both

$bc + 20 = \sqrt{3 } bc \\ \sqrt{3} bc - bc = 20 \\ bc( \sqrt{3} - 1) = 20 \\ bc = \frac{20}{ \sqrt{3 } - 1}$

$and \: ab = \sqrt{3} bc \\ so \: ab = \frac{20 \sqrt{3} }{ \sqrt{3} - 1}$

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